# How to add days to date in C++

In this program, we will add a given number of days to the given date in C++. Here we will display the date after adding the days as given in input. We have different cases for adding the number of days to the given date. We will calculate the total number of days in the given date and then add number of days to be added to those days. Now, these total days are used to calculate the new date after adding the number of days.

First thing is to calculate the total number of days on a given date. Let us consider d be the number of days in the given date, be the month of the given date and  be the year on the given date. Consider ne to be the number of days to be added to the given date. Now the source code to get the total number of days in given date is:

```int m2[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int i,k=0;
for(i=0;i<m;i++)
k=k+m2[i];
k=k+d+ne;```

From the above source code  is the total number of days on the given date.

Now we need to add number of days we want and display the new date. For that, we have two cases if the total days after adding the number of days ne is greater than 365 than add 1 year to the given date or else calculate the date from total days and print the year the same as on the given date.

## Code for adding number of days we want to add to given date in C++

```#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
int d,m,y;
cout<<"\n enter days=";
cin>>d;
cout<<"\n enter month=";
cin>>m;
cout<<"\n enter year=";
cin>>y;
int ne;
cout<<"\n enter number of days you want to add=";
cin>>ne;
int m2[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int i,k=0,p,a;
for(i=0;i<m;i++)
k=k+m2[i];
k=k+d+ne;
cout<<\n new date=";
if(k<=365)
{
for(i=0;i<13;i++)
{
p=k-m2[i];
if(p<=m2[i+1])
{
a=i+1;
break;
}
else
k=p;
}
cout<<p<<"/"<<a<<"/"<<y;
}
else
{
k=k-365;
for(i=0;i<13;i++)
{
p=k-m2[i];
if(p<=m2[i+1])
{
a=i+1;
break;
}
else
k=p;
}
cout<<p<<"/"<<a<<"/"<<y+1;
}
getch();
}

```

Output:

```enter days=12
enter month=5
enter year=2019
enter number of days you want to add=35
new date=16/6/2019```

In the given approach, we use rational brute force method to add given number of days to a given date in C++.

Firstly we write smaller functions to be used in our program like. isleapyear is used to check if a year is a leap or not. offsetdays to calculate offset i.e. the number of days elapsed from the beginning of the year(1st january). adddays to add the given number of days in our entered date. revoffsetdays for given a year and number of days passed in it, finds date by storing resulting data into d and m.

In the driver code, we can either ask the user to enter input for date and the number of days to be added or we can directly put that in main function.

## Add days to date in C++ – Second method

The given is a programming approach for the same.

```bool isleapyear(int year)
{
if (year % 400 == 0)
return true;

if (year % 100 == 0)
return false;

if (year % 4 == 0)
return true;
return false;
}```
```int offsetdays(int d, int m, int y)
{
int offset = d;

switch(m-1)
{
case 1:
offset=offset+ 31;
case 2:
offset=offset+ 28;
case 3:
offset=offset+ 31;
case 4:
offset=offset+ 30;
case 5:
offset=offset+ 31;
case 6:
offset=offset+ 30;
case 7:
offset=offset+ 31;
case 8:
offset=offset+ 31;
case 9:
offset=offset+ 30;
case 10:
offset=offset+ 31;
case 11:
offset=offset+ 30;

}

if (isleapyear(year) && m > 2)
offset += 1;

return offset;
}```
```int montharray = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
void revoffsetdays(int offset, int year, int *d, int *m)
{
if (isleapyear(year))
{montharray = 29; }
integer i;
for (i = 1; i <= 12; i++)
{
if (offset <= montharray[i])
break;

offset = offset - montharray[i];
}
*d = offset;
*m = i;
}

void addDays(int d1, int m1, int year1, int x)
{ int offset1 = offsetdays(d1, m1, year1);
int remdays = isleapyear(year1)?(366-offset1):(365-offset1);
// year2 stores result year
// offset2 stores the number of offset days in result year.
int year2, offset2;
if (x <= remdays)
{ year2 = year1; offset2 = offset1 + x; }

else
{
x -= remdays;
year2 = year1 + 1;
int year2days = isleapyear(year2)?366:365;
while (x >= y2days) { x -= y2days; y2++; y2days = isleapyear(y2)?366:365; }
offset2 = x;
}
int m2, d2;
revoffsetdays(offset2, year2, &d2, &m2);
cout <<d2<<":"<< m2 << ":" r; }

int main()
{ int d=21, m=2, y=2019;
int x = 369;
`25:2:2020`